Question

与No argument names in abstract declaration?相关，您如何使用C＃样式的tupled参数执行此操作？所以，如果我想要

abstract member createEmployee : (string * string) -> Employee

我怎么能沟通是否应该

1-   member this.createEmployee(firstName, lastName) = ...
2-   member this.createEmployee(lastName, firstName) = ...

使用案例：在F＃中创建一个用于从C＃中使用的接口。

Answer 1

abstract createEmployee: firstName:string * lastName:string -> Employee

编译以创建C＃调用语义，并且在Visual Studio 2013中，名称同时出现在C＃和F＃intellisense中。

Answer 2

我认为最简单的方法就是这样：

type FirstName = FirstName of string
type LastName  = LastName of string

...

abstract member createEmployee : (FirstName * LastName) -> Employee

备注：

你可能希望它是

abstract member createEmployee : FirstName * LastName -> Employee

因为存在细微差别;）

Answer 3

有两种方法可以做到：

使用MSDN

中所述的命名参数

member this.createEmployee(firstName: string, lastName: string) = ...

// external code
MyClass.createEmployee(firstName = "John", lastName = "Smith")
MyClass.createEmployee(lastName = "Smith", firstName = "John")

让您的类成员接受一个结构作为参数：

type EmployeeName =
   struct 
      val firstName: string
      val lastName: string
   end 

member this.createEmployee2(employeeName: EmployeeName) = ...

// external code
MyClass.createEmployee2 {firstName = "John"; lastName = "Smith"}
MyClass.createEmployee2 {lastName = "Smith"; firstName = "John"}

选择权归你的;前者允许使用可选参数（考虑middleName或salutation），而后者允许将结构存储在一起（例如，如果您需要进一步处理它）。

抽象tupled函数中的参数名称

3 个答案:

备注：